Me Todos
-
Upload
bebeto-sotelo -
Category
Documents
-
view
218 -
download
0
description
Transcript of Me Todos
![Page 1: Me Todos](https://reader036.fdocuments.mx/reader036/viewer/2022082715/5695cfbb1a28ab9b028f4e45/html5/thumbnails/1.jpg)
11. En cada una de las siguientes ecuaciones determine un intervalo (a,b) en que convergerá la iteración. Estime la cantidad de iteraciones necesarias para obtener aproximadamente con una exactitud de 10 -5 y realice los cálculos.
a. x=2−ex+x2
3Solución:
x2−ex−3x+2=0f x=x
2−ex−3 x+2
i. METODO DE NEWTON f x=x
2−ex−3 x+2f ' x=2 x−e
x−3
xn+1=xn−xn2−e xn−3 xn+22 xn−e
xn−3
n xn xn−xn+10 11 0.268941421 0.731058582 0.434457296 0.165515873 0.447988829 0.013531534 0.448063074 7.4245x10-5
ii. METODO DE LA SECANTE f x=x
2−ex−3 x+2
xn+1=xn−(xn−1−xn)(xn
2−exn−3 xn+2)xn−1
2−exn−1−3 xn−1−(xn2−exn−3 xn)
n xn xn−xn+10 01 1 12 0.26894142 0.731058583 0.25717072 0.01177074 0.25753067 0.000359945 0.25753029 3.8123x10-7
∴La iteracion convergeenel intervalo(0,1)
![Page 2: Me Todos](https://reader036.fdocuments.mx/reader036/viewer/2022082715/5695cfbb1a28ab9b028f4e45/html5/thumbnails/2.jpg)
b. x=5x2
+2
Solución:x3−5+2 x3=0f x=x
3+2 x3−5
i. METODO DE NEWTON f x=x
3+2 x3−5f ' x=3 x
2−4 x
xn+1=xn−xn3+2 xn
3−53 xn
2−4 xn
n xn xn−xn+10 31 2.733333333 0.2666666672 2.691624726 0.0417086083 2.690647977 0.0009767494 2.690647448 5.28901E-07
ii. METODO DE LA SECANTE f x=x
3+2 x3−5
xn+1=xn−(xn−1−xn)(xn
3+2 xn3−5)
xn−13+2xn−1
3−(xn3+2 xn
3)
n xn xn−xn+10 21 3 12 2.55555556 0.444444443 2.66905005 0.11349454 2.69236876 0.023318715 2.69062668 0.00174208
2.69064743 2.0744E-05
∴La iteracion convergeenel intervalo(2,3)
![Page 3: Me Todos](https://reader036.fdocuments.mx/reader036/viewer/2022082715/5695cfbb1a28ab9b028f4e45/html5/thumbnails/3.jpg)
c. x=√ ex33 x2−ex=0f x=3 x
2−e x
i. METODO DE NEWTON f x=3 x
2−e x
f ' x=6x−ex
xn+1=xn−3 xn
2−exn
6 xn−exn
n xn xn−xn+10 11 0.914155282 0.0858447182 0.910017666 0.0041376163 0.910007572 1.00932E-05
ii. METODO DE LA SECANTE f x=3 x
2−e x
xn+1=xn−(xn−1−xn)(3 xn
2−exn)3 xn−1
2−exn−1−(3xn2−exn)
∴La iteracion convergeenel intervalo(0,1)
![Page 4: Me Todos](https://reader036.fdocuments.mx/reader036/viewer/2022082715/5695cfbb1a28ab9b028f4e45/html5/thumbnails/4.jpg)
d. x=5−x
x−5−x=0f x=x−5
− x ln (5)i. METODO DE NEWTON
f x=x−5− x
f ' x=1−5− x
xn+1=xn−x−5− x
1−5− x ln (5)
n xn xn−xn+10 11 0.394804815 0.605195182 0.467629704 0.072824893 0.469620547 0.001990844 0.469621923 1.3759E-06
ii. METODO DE LA SECANTE f x=x−5
− x
xn+1=xn−(xn−1−xn)(xn−5
−xn)
xn−1−5−xn−1−(xn−5
− xn)
n xn xn−xn+10 01 1 12 0.55555556 0.444444443 0.4558435 0.099712054 0.4700264 0.01418295 0.46962386 0.00040254
0.46962192 1.9357E-06
∴La iteracion convergeenel intervalo(0,1)
![Page 5: Me Todos](https://reader036.fdocuments.mx/reader036/viewer/2022082715/5695cfbb1a28ab9b028f4e45/html5/thumbnails/5.jpg)
e. x=6−x
x−6−x=0f x=x−6
− x ln (6)i. METODO DE NEWTON
f x=x−6− x
f ' x=1−6−x
xn+1=xn−x−6−x
1−6− x ln (6)
n xn xn−xn+10 11 0.358296413 0.6417035872 0.444740128 0.0864437153 0.448058666 0.0033185374 0.448063077 4.41096E-06
ii. METODO DE LA SECANTE f x=x−6
− x
xn+1=xn−(xn−1−xn)(xn−6
− xn)
xn−1−6−xn−1−(xn−6
−xn)
n xn xn−xn+10 01 1 12 0.54545455 0.454545453 0.4297045 0.115750044 0.44876504 0.019060545 0.44806824 0.00069686 0.44806308 5.1606E-06
∴La iteracion convergeenel intervalo(0,1)